Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

Answer 1

We start by proving the statement is true for $n=1$. In this case, the left hand side is simply:

$(3\times 1-2)^2 = 1.$

The right hand side is:

$\dfrac{1\times (6 \times 1^2 - 3\times 1 -1)}{2} = \dfrac{2}{2}=1.$

So the property holds for $n=1$.

Now we assume the property is valid for some $n$ and prove that it implies that it is also valid for $n+1$. That means we shall obtain:

$1^2 + 4^2 + 7^2 + \dots + (3n-2)^2 + [3(n+1)-2]^2 = \dfrac{(n+1)[6(n+1)^2-3(n+1)-1]}{2}.$

We start by writing the property for $n+1$ and recognize that we've only added an extra term:

$\underbrace{1^2 + 4^2 + 7^2 + \dots + (3n-2)^2}_{=\dfrac{n(6n^2-3n-1)}{2}} + [3(n+1)-2]^2.$

We now expand the square and merge everything together in a single fraction:

$\dfrac{n(6n^2-3n-1)}{2} + [3(n+1)-2]^2 = \dfrac{6n^3-3n^2-n}{2} + (3n+1)^2 = \\\\= \dfrac{6n^3-3n^2-n}{2} + (9n^2+6n+1) = \dfrac{6n^3 - 3n^2 -n + 18n^2+12n+2}{2} =\\\\= \dfrac{6n^3+15n^2+11n+2}{2}.$

Since the desired expression has a $n+1$ facto, we may now divide the numerator by $n+1$, using, for instance, Ruffini's rule, in order to find it:

$\begin{array}{c|ccc|c} & 6 & 15 & 11 & 2\\ -1 & &-6 & -9 & -2 \\ -&-&-&-&-\\ & 6 & 9&2&0\end{array}$

So we get:

$\dfrac{6n^3+15n^2+11n+2}{2} = \dfrac{(n+1)(6n^2+9n+2)}{2}.$

We now add and subtract $12n$ and $6$ in order to obtain the form corresponding to $(n+1)^2$, since we know that it must appear:

$6n^2+9n+2 = 6n^2 + 12n + 6 - 12n - 6 + 9n+2 =\\\\=6\underbrace{(n^2+2n+1)}_{=(n+1)^2}-3n-4 = 6(n+1)^2 - 3n - 3 -1 = 6(n+1)^2 - 3(n+1)-1.$

So we finally got:

$1^2 + 4^2 + 7^2 + \dots + (3n-2)^2 + [3(n+1)-2]^2 = \dfrac{(n+1)[6(n+1)^2-3(n+1)-1]}{2},$

which is the desired result. So we proved that the statement is true for every integer $n \geq 1$.